Find the derivative of LaTeX: \displaystyle y = \frac{\left(x - 1\right)^{5} \left(6 x + 4\right)^{6} e^{- x}}{\left(- 5 x - 5\right)^{7} \cos^{8}{\left(x \right)}}

Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: LaTeX: \ln(y) = \ln{\left(\frac{\left(x - 1\right)^{5} \left(6 x + 4\right)^{6} e^{- x}}{\left(- 5 x - 5\right)^{7} \cos^{8}{\left(x \right)}} \right)} Expanding the right hand side using the product and quotient properties of logarithms gives: LaTeX: \ln(y) = 5 \ln{\left(x - 1 \right)} + 6 \ln{\left(6 x + 4 \right)}- x - 7 \ln{\left(- 5 x - 5 \right)} - 8 \ln{\left(\cos{\left(x \right)} \right)} Taking the derivative on both sides of the equation yields: LaTeX: \frac{y'}{y} = \frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{36}{6 x + 4} + \frac{5}{x - 1} + \frac{35}{- 5 x - 5} Solving for LaTeX: \displaystyle y' and substituting out y using the original equation gives LaTeX: y' = \left(\frac{8 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{36}{6 x + 4} + \frac{5}{x - 1} + \frac{35}{- 5 x - 5}\right)\left(\frac{\left(x - 1\right)^{5} \left(6 x + 4\right)^{6} e^{- x}}{\left(- 5 x - 5\right)^{7} \cos^{8}{\left(x \right)}} \right) Using some Trigonometric identities to simplify gives LaTeX: y' = \left(\frac{36}{6 x + 4} + \frac{5}{x - 1}8 \tan{\left(x \right)} - 1 + \frac{35}{- 5 x - 5}\right)\left(\frac{\left(x - 1\right)^{5} \left(6 x + 4\right)^{6} e^{- x}}{\left(- 5 x - 5\right)^{7} \cos^{8}{\left(x \right)}} \right)